As a student in actuarial science, I have :
Consider a compound Poisson model for the aggregate claim size S in an insurance portfolio with 60 expected claims per time unit and each claim following a Pareto distribution with x0 = 12 and α = 2.5. a) Determine the standard deviation of S
Is it right to get it with the moment generating function of pareto including (e^(t)-1) instead of t ?
Recall the law of total variance: if $S = X_1 + X_2 + \cdots + X_N$ where $$X_i \sim X \sim \operatorname{Pareto}(x_0 = 12, \alpha = 2.5), \quad f_{X}(x) = \begin{cases} \frac{x_0^\alpha}{\alpha x^{\alpha+1}}, & x \ge x_0 \\ 0, & \text{otherwise},\end{cases}$$ and $$N \sim \operatorname{Poisson}(\lambda = 60),$$ then we have $$\begin{align*} \operatorname{Var}[S] &= \operatorname{Var}[\operatorname{E}[S \mid N]] + \operatorname{E}[\operatorname{Var}[S \mid N]] \\ &= \operatorname{Var}[N \operatorname{E}[X]] + \operatorname{E}[N \operatorname{Var}[X]] \\ &= \operatorname{E}[X]^2 \operatorname{Var}[N] + \operatorname{Var}[X] \operatorname{E}[N]. \end{align*}$$ This formula should be familiar to you; if not, you should memorize it.
In the case of the Poisson distribution, $\operatorname{E}[N] = \operatorname{Var}[N] = \lambda$. In the Pareto distribution, we have $$\operatorname{E}[X] = \frac{\alpha x_0}{\alpha-1}, \quad \operatorname{Var}[X] = \frac{\alpha x_0^2}{(\alpha-1)^2 (\alpha-2)},$$ hence $$\operatorname{Var}[S] = \frac{\lambda \alpha x_0^2}{\alpha - 2} \quad \alpha > 2.$$